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# The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle

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Math Expert
Joined: 02 Sep 2009
Posts: 56303
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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26 Mar 2015, 04:36
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Difficulty:

55% (hard)

Question Stats:

70% (02:50) correct 30% (02:53) wrong based on 159 sessions

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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Kudos for a correct solution.

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Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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30 Mar 2015, 03:37
4
1
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:
Attachment:

righttrianglearea_text.PNG [ 17.63 KiB | Viewed 8484 times ]

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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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26 Mar 2015, 06:01
4
1
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Kudos for a correct solution.

B and C should have same Y co-ordinate --> $$4a-5 = 2a+6$$ --> a=$$\frac{11}{2}$$

area of triangle = $$\frac{1}{2} * (4a-5) * (2a+1)$$ . on substituting a=$$\frac{11}{2}$$
we get area = 102 .

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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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26 Mar 2015, 07:14
2
1
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Kudos for a correct solution.

1/2bh=1/2(2a+1)(2a+6)

Now 4a-5=2a+6
2a=11

Therefore,
A(0,0); B(0,17); C(12,17)

1/2*17*12=102

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Joined: 05 Aug 2016
Posts: 1
Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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12 Jan 2018, 12:22
AmoyV wrote:
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

Now 4a-5=2a+6

Sorry, but I did not get to your calculations - why it is so? What I missed?
Intern
Joined: 12 Dec 2017
Posts: 3
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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19 Jan 2018, 02:13
1
Aksena wrote:
AmoyV wrote:
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

Now 4a-5=2a+6

Sorry, but I did not get to your calculations - why it is so? What I missed?

Im guessing you have problems understanding why 4a-5 = 2a+6

Now, the reason they got to that equation was due to the given information that angle ABC = 90
--> This implies that triangle ABC was a right angle
--> Now A was given at the origin (0,0), while B was given to be somewhere along the Y-axis as its X-coordinates = 0
--> Hence, as ABC was a right triangle, the point C will have the same Y-coordinates as B in order to form the right triangle (i.e. to fulfill angle ABC = 90)
--> B(0,4a-5) and C(2a+1, 2a+6)
--> Therefore 4a-5 = 2a+6, solve for a and you can find the area of the triangle with the basic area formula for right triangles: base x height / 2
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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21 Jan 2018, 19:33
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Since we have a right triangle with angle ABC = 90, we see that vertices A and B share the same x-coordinate, and vertices B and C share the same y-coordinate.

Equating the y-coordinates of vertices B and C, we have:

4a - 5 = 2a + 6

2a = 11

a = 5.5

Substituting 5.5 for a, we see that the ordered pair for vertex B is (0, 4(5.5) - 5), or (0, 17). Thus, leg AB is the vertical distance from (0,0) to (0,17), or 17.
Similarly, we see that the ordered pair for vertex C is (12, 17) and leg BC is is the horizontal distance from (0,17) to (12, 17) = 12.

Thus, the area of the triangle is:

area = base x height x 1/2

area = 17 x 12 x 1/2 = 17 x 6 = 102

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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle   [#permalink] 09 Mar 2019, 01:14
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